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I would like to use numerical integration to approximate $\int_{0}^{1} f(x) dx$ where $f(x) = \frac{1}{\sqrt{x}}$. But I can't figure out how to get an error bound.

For example, if I use trapezoidal rule then the error bound is

$$|E|\leq K\frac{1}{12\cdot n^2}$$

where $|f''(x)|\leq K$. But in this case there is no $K$ since $|f''(0)|$ is infinite.

I have a similar issue if I try to use Simpson's rule.

Is there a numerical integration method where I could calculate a error bound for this problem?

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    $\begingroup$ Your function $f$ is not defined at $0$ so applying the trapezoidal rule directly is not going to work. Would you be interested in the case $g(x) = x^\frac{1}{2}$ for which the trapezoidal rule is defined and where the standard error bound does not apply either? $\endgroup$ – Carl Christian Feb 1 '16 at 15:41
  • $\begingroup$ I picked this $f$ because even though it is not defined at 0, the integral between 0 and 1 converges. I'm curious if there is a way to calculate an error bound for this type of functions. $\endgroup$ – Daniel Vaca Feb 1 '16 at 17:35
  • $\begingroup$ The following paper derives both the classical and some less common error estimates for the trapzoidal rule using partial integration. maa.org/sites/default/files/An_Elementary_Proof30705.pdf The authors assume slightly more than you allow, but it is still worth a read. $\endgroup$ – Carl Christian Feb 1 '16 at 19:07
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The simplest approximation that works for this improper integral is the lower Riemann sum where

$$E = \left| \frac{1}{n} \sum_{k=1}^{n} \frac{1}{\sqrt{k/n}} - \int_0^1 \frac{dx}{\sqrt{x}} \right|= \left|\frac{1}{\sqrt{n}} \sum_{k=1}^{n} \frac{1}{\sqrt{k}} - 2 \right| < O(1/ \sqrt{n}).$$

The RHS error bound is obtained by summing with the inequalities

$$ \frac{1}{\sqrt{k}} > \frac{2}{\sqrt{k} + \sqrt{k+1}} = 2( \sqrt{k+1} - \sqrt{k}) \\ \frac{1}{\sqrt{k}} < \frac{2}{\sqrt{k} + \sqrt{k-1}} = 2( \sqrt{k} - \sqrt{k-1}),$$

and using a Taylor expansion.

Specifically,

$$\frac{1}{\sqrt{n}} \sum_{k=1}^{n}2( \sqrt{k+1} - \sqrt{k}) < \frac{1}{\sqrt{n}} \sum_{k=1}^{n} \frac{1}{\sqrt{k}} < \frac{1}{\sqrt{n}} \sum_{k=1}^{n}2( \sqrt{k} - \sqrt{k-1}), $$

whence,

$$ \sqrt{1 + 1/n} - 1/\sqrt{n} < \frac{1}{\sqrt{n}} \sum_{k=1}^{n} \frac{1}{\sqrt{k}} < 2, $$

and

$$E < 2(1 + 1/\sqrt{n} - \sqrt{1+1/n}) = O(1 / \sqrt{n}).$$

In general for such improper integrals, it is advisable to try to remove the singularity before applying a numerical approximation. In some cases, you can simply ignore the singularity and apply the trapezoidal or Simpson rule with the value $f(0)$ arbitrarily set to $0$, or use something like Gaussian quadrature that does not explicitly involve the value $f(0)$. However, the convergence rate will be degraded and most certainly will not be $O(1/n^2)$. Furthermore, this may not work well if the integrand oscillates, e.g. $f(x) = (1/x) \sin (1/x).$

If the integrand is of the form $x^{-1/2}f(x)$, where $f \in C([0,1])$, then you can apply the transformation $s^2 = x$ and obtain a proper integral that is amenable to numerical approximation:

$$\int_0^1 x^{-1/2}f(x) \, dx = 2\int_0^1 f(s^2) \, ds.$$

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